Witten’s Problems, Set Four — N^{o}. 2

(solution written by D. Freed)

For practice, we solve this problem using dimensional regularization. The dimensions of the fields and parameters are

[ϕ] = \frac{n−2}{2}
[m] = 1
[g] = \frac{6−n}{2}

Introduce a parameter μ with the dimension of mass and a dimensionless parameter λ. Then the lagrangian density is

ℒ = \{\frac{1}{2} (|dϕ|^{2} + m^{2}ϕ^{2}) + λμ^{\frac{6−n}{2}} \frac{ϕ^{3}}{3!} \} d^{n}x .

To order λ^{3} the 3-point function is computed by the diagram shown. Here we are working in

.^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} p_{1}  p_{2} q+p_{1} q  −(p_{1}+p_{2}) q+p_{1}+p_{2}

momentum space and label the edges of the diagram with momenta. Applying the Feynman rules for Euclidean space, we evaluate the diagram as

Γ = −λ^{3}μ^{\frac{3}{2} (6−n)} ∫ \frac{d^{n}q}{(2π)^{n}} \frac{1}{(q^{2}+m^{2})((q+p_{1})^{2}+m^{2})((q+p_{1}+p_{2})^{2}+m^{2})} Now use Feynman parameters to rewrite the integral as

(1)  Γ_{n} = −λ^{3}μ^{\frac{3}{2} (6−n)} ∫ \frac{d^{n}q}{(2π)^{n}} ∫_{0}^{1} dα ∫_{0}^{1−α} dβ \frac{1}{[(1−α−β)q^{2}+α(q+p_{1})^{2}+β(q+p_{1}+p_{2})^{2}+m^{2}]^{3}}
= −λ^{3}μ^{\frac{3}{2} (6−n)} ∫_{0}^{1} dα ∫_{0}^{1−α} dβ ∫ \frac{d^{n}q}{(2π)^{n}} \frac{1}{[q^{2}+A^{2}]^{3}} ,

where

A^{2} = m^{2} + α(1 − α)p_{1}^{2} + β(1 − β)(p_{1} + p_{2})^{2} − 2αβp_{1}(p_{1} + p_{2}). (2)

the interchange of integrals is justified for n sufficiently small (here n < 6). Note that we made a shift in the integration variable q. Now the rules for integrals in n dimensions (n complex!) give

\begin{aligned}∫ \frac{d^{n}q}{(2π)^{n}} \frac{1}{[q^{2}+a^{2}]^{3}} & = \frac{1}{(2π)^{n}} \frac{π^{n/2}}{\frac{1}{2} Γ(n/2)} ∫_{0}^{∞} dr r^{n−1} \frac{1}{[r^{2}+A^{2}]^{3}} \\ & = \frac{1}{2^{n−1}π^{n/2}Γ(n/2)} · \frac{1}{2} \frac{Γ(\frac{n}{2})Γ(3− \frac{n}{2})}{Γ(3)}A^{n−6} \\ & = \frac{Γ(3− \frac{n}{2})}{2^{n+1}π^{n/2}} A^{n−6}. \end{aligned} (3)

Here \frac{π^{n/2}}{\frac{1}{2} Γ(n/2)} is the volume of the unit sphere in n dimensions, and for any N,

∫_{0}^{∞} dr r^{n−1} \frac{1}{[r^{2}+A^{2}]^{N}} = ∫_{0}^{∞} dr r^{n−1} \frac{1}{Γ(N)} ∫_{0}^{∞} dt e^{−(r^{2}+A^{2})t}t^{N−1}
= \frac{A^{n−2N}}{Γ(N)} ∫_{0}^{∞} \frac{dt}{t} ∫_{0}^{∞} dr e^{−r^{2}t}r^{n−l}e^{−t}t^{N}
= \frac{A^{n−2N}}{2Γ(N)} ∫_{0}^{∞} \frac{dt}{t} ∫_{0}^{∞} \frac{du}{u}e^{−u}u^{\frac{n}{2}} e^{−t}t^{N−n/2}
= \frac{1}{2} \frac{Γ(\frac{n}{2})Γ(N−n/2)}{Γ(N)}A^{n−2N} .

The interchange of integrations is valid for n small. Now (3) is finite for n < 6 and has a simple pole at n = 6. Collecting (1) and (3), we find

\begin{aligned}Γ_{n} & = −\frac{λ^{3}}{2^{n+1}π^{n/2}} Γ (\frac{6−n}{2}) μ^{\frac{3}{2} (6−n)} ∫_{0}^{1} dα ∫_{0}^{1−α} dβA^{n−6} \\ & = −\frac{λ^{3}B_{n}}{2^{n+1}π^{n/2}} μ^{\frac{3}{2} (6−n)}Γ (\frac{6−n}{2}) , \end{aligned}

where

B_{n} = ∫_{0}^{1} dα ∫_{0}^{1−α} dβ [m^{2} + α(1 − α)p_{1}^{2} + β(1 − β)(p_{1} + p_{2})^{2} − 2αβp_{1}(p_{1} + p_{2})]^{\frac{n−6}{2}} .

Note B_{6} = 1/2 and B_{n} is a holomorphic function of n near n = 6. Introduce ε = 3 − n/2 and write

Γ_{n} = −λ^{3}μ^{ε} \{\frac{1}{2(4π)^{3}} Γ(ε)(4πμ^{2})^{ε}B_{n}\}

and then expand around ε = 0:

\begin{matrix}Γ_{n} = \\ − λ^{3}μ^{ε}.\frac{1}{2(4π)^{3}} \{\frac{1}{2ε} − \frac{γ}{2} − ∫_{0}^{1}dα ∫_{0}^{1−α}dβ ln (\frac{m^{2}+α(1−α)p_{1}^{2}+β(1−β)(p_{1}+p_{2})^{2}−2αβp_{1}(p_{1}+p_{2})}{4πμ^{2}})\} + O(ε), \end{matrix}

where γ is Euler’s constant.

To renormalize for n = 6, we introduce the new lagrangian density

ℒ′ = \{\frac{1}{2} (|dϕ|^{2} + m^{2}ϕ^{2}) + λ (1 + \frac{λ^{2}}{2(4π)^{3}} \frac{1}{6−n}) μ^{\frac{6−n}{2}} \frac{ϕ^{3}}{3!} \} d^{n}x .

This introduces a diagram which cancels the pole in Γ_{n} at n = 6, and the new 3-point function to one loop is the finite part of Γ_{n} at n = 6:

\frac{λ^{3}}{2(4π)^{3}} \{\frac{γ}{2} + ∫_{0}^{1} dα ∫_{0}^{1−α} dβln (\frac{m^{2}+α(1−α)p_{1}^{2}+β(1−β)(p_{1}+p_{2})^{2}−2αβp_{1}(p_{1}+p_{2})}{4πμ^{2}})\} .